15- Solved problem 6-19-4 for local buckling of column-2010.

Last Updated on July 26, 2025 by Maged kamel

Solved problem 6-19-4 for local buckling of column-2010.

Solved problem 6-19-4-estimate area and inertia values of the given section.

The next slide image shows a brief introduction to the content of this post..

 What is the main chapter for the design of compression members? The main chapter is Chapter E, and the relevant sections for the built-up section are E6 and E7 for slender members.

 Determine the nominal axial strength for the Non-standard shape of Figure 6.19.4 for an effective length kl=8 feet, with fy used of 100 ksi. The I  section is composed of three plates; the upper plate breadth is 10 inches. The thickness is 1/2 inches.

The same dimension is also for the lower plate. The middle plate has a height =11 inches, with a width of 1/4 inch; thus, the overall height is 12 inches.
Given k*L is 8 feet. Since this section is a built-up section,  we can not get information from any tables. The following slide image shows the calculation of Ix, or the moment of inertia about x-x, the Ix value equals 358.56 inch4.

For local buckling, we have to estimate Ix and Iy. The value of rx is much higher than ry, the minor direction will govern buckling.

We will estimate the Iy, since buckling will be about the minor axis. the area of section equals 12.75 inch2, the moment of inertia about Y axis equals 83.35 inch4. The radius of gyration equals 2.56 inches. Please refer to the following slide image for more details.

Determine which direction controls the design.

We can find the value of the radius of gyration about the x-x, which is sqrt(358.56/12.75)=5.30 inches. Find the values of (Kl/rx) and Kl/ry, For Kl/rx=96/5.3=11.62.

While Kl/ry=(96/2.56)=37.50. Since Kl/ry is bigger than Kl/rx, the buckling about Y-axis will control the design.

Check b/t for unstiffened part-Flange and reduction factor Qs for unstiffened part.

The ratio of b/t is the ratio of half of the flange length divided by flange, upper plate thickness, equals (10/2/0.50)=10. Check this ratio with B/t as given by E7-7 Equation, which gives this relation, b/t <=0.64*sqrt(E*kc/Fy).

The kc value equals 4/sqrt(h/tw)=4/sqrt(11/4)=0.603, since hw=11″ and web thickness, middle plate=0.25 inches. The required b/t =0.64*sqrt(29000*0.603/100)=8.46.

Since b/t of the section is bigger than 8.46, Qs is not equal to one. We need to proceed to E7-8 equation to find the equation for Qs.

Estimate b/t based on the value of 1.17*sqrt(E*kc/Fy), which equals(1.17*sqrt(29000*0.603/100)=15.47.

The actual b/t,10, is bigger than 8.46 but less than 15.47. The Qs equation 1.145-0.65*(b/t)*sqrt(Fy/E*kc)=1.145-10*sqrt(100/29000*0.603)=0.923. Please refer to the following slide image.

For the stiffened part, the ratio of (h/tw)=11/0.25=44.

Find the affection b value and the effective area based on Q=1.0.

Now, we will proceed to the calculation of  Qa. For the slender stiffened element, h/tw=11/0.25=44 will be compared with (1.49*sqrt(E/Fy),  which gives the value of  1.49*sqrt(29000/100)=25.40.

Our h/tw for the section > 25.4, since h/tw>λr, the Qa is <1, since h/tw is >λr. Then the modified h will be based on E7-17. But we need to find the value of stress f in the equation, consider Qa=1.0.

Find the Euler stress and critical stress for the column.

The column is inelastic since Kl/r y value which equals 37.50, is less than 4.71*sqrt(E/Q*Fy) which equals 80.208, The facr=0.658^(Q*fy/Fe)*(Q*Fy). The Euler stress=Pi^2*E/(Kl/r)^=204 ksi,

First, consider Qa=1, our calculated Qs=0.923, then q=1*0.923=0.923,for the first trial, Qfy=0.923*100=92.3 ksi, Fcr=Qfy*(0.658^Qfy/fE)=92.3* 0.658^(92.3/204)=76.38 ksi.

The new graph for the short column is shown as a dotted line.
Take the fcr value, which is =76.38 ksi, back to the equation E 7-17 as follows.

Find the effective b-value and the effective area based on Q=0.923.

Then be=1.92*(tw)*sqrt (E/f)*(1-(0.34/bt*Sqrt(E/f), tw=0.25 *sqrt(29000/f), the trial f=Qa*Fcr as estimated earlier, the trial f=0.923*76.28, Trial f=0.923*76.38=70.50 ksi, the term 1-(0.34/bt*Sqrt(E/f)=1-(0.34/11/0.25*sqrt(29000/70.5)=8.209 inches<11 inches which is the web height.

The effective area=Ag-hw*hw+be*tw=12.75-11*0.25+8.209*0.25=12.05 inch2. Next, we will estimate the Qa value.

Reestimate the Qa=Aeff /Agr=12.05/12.75=0.945. For the new value of Qa. Qa=12.05/12.75=0.9452. So the Q new=Qs*Qa, the value of Qa=0.945 while Qs=0.923.

Finally, Q new=0.923*0.945=0.8722, back to Fcr =72.94 ksi. Again, multiply by Q = 0.8722 to get another value of stress f=0.8722*72.94=63.60 ksi.

Multiply by Q to get the new f, which is 0.8722*72.9=63.6 ksi.

Find the affection b value and the effective area based on Q=0.8722.

Evaluate the modified value of (be ) for the stiffened web, be=1.92*(0.25)*sqrt(29000/63.6)* by the other bracket—finally, hw eff=8.557 inches.

Recalculate the A eff=10*0.5*2+8.557*0.25, Aeff =12.14 inch2. The corresponding effective area is equal to 12.14 inch2.

Find the effective b-value and the effective area based on Q=0.872.

Qa new =A eff/Ag=12.14/12.75=0.945, To get the  Q=Qs*Qa=0.923*0.945= 0.872,  fcr recalculated with the new Q value, Fcr=72.94 ksi. Multiply by Q to get the new f, which is 0.872*72.94=63.6 ksi. Evaluate the modified value of (be ) for the stiffened web, be=1.92*(0.25)*sqrt(29000/63.6)* by the other bracket—finally, hw eff=8.557 inches, same as estimated earlier.

Solved problem 6-19-4-estimate the nominal load.

The A eff=12.75-(11*0.25)+(8.557*0.25)=12.14 inch2. The Final Qa value=12.14/12.75=0.952. The final Q value=0.923*0.952=0.879. Estimate fcr=0.658^(0.879*100/204)*87.90=73.0 ksi. there Nominal load =fcr *Ag=73*12.75=931 kips.

In this example for LRFD only, accordingly, φc, φc*Pn =0.9*931=838 kips, this is the end of our solved problem 6-19-4. Thanks a lot.

For a useful external resource,  please review A Beginner’s Guide to Structural Engineering.
For the next post, A Solved Problem 5-10 For the Available Strength.